Question

Another model for a growth function for a limited population is given by the Gompertz function, which is a solution to the differential equation dPdt=cln(KP)P where c is a constant and K is the carrying capacity. Answer the following questions. 1. Solve the differential equation with a constant c=0.05, carrying capacity K=3000, and initial population P0=1000.

Answer 1

Answer:

The solution is

$P(t) = 1000e^{-1}e^{e^{-0.05t}}$.

Step-by-step explanation:

We have the Gompertz's differential equation

$\frac{dP}{dt} = -cP\ln(KP)$.

Notice that this a separable equation. So,

$\frac{dP}{P\ln(KP)} = -cdt$.

Now, integrating in both sides:

$\int\frac{dP}{P\ln(KP)} = -\int cdt$.

Recall that

$\int\frac{dP}{P\ln(KP)} = \ln(|\ln(KP)|)$.

Thus,

$\ln(|\ln(KP)|) = -ct+A$,

and taking exponential:

$|\ln(KP)| = A'e^{-ct}$.

Taking another exponential

$P(t) = \frac{A''}{K} e^{e^{-ct}}$.

Now, let us substitute the given values for c and K:

$P(t) = \frac{A''}{3000} e^{e^{-0.05t}}$.

To find the value of the constant A'' we use the initial condition:

$P(0) = \frac{A''}{3000} e^{e^{0}} = \frac{A''}{3000}e=1000$.

Then,

$A'' = \frac{3\cdot 10^6}{e}$.

Hence, the solution is

$P(t) = 1000e^{-1}e^{e^{-0.05t}}$